text:
Are zero visibility landings a possibility for the future of commercial aviation?
Airports usually close or slow down activities in fog conditions, e.g. this seems to be a significant problem in winter at Delhi airport.
Large commercial aircraft would be able to land with no visibility (more accurately with no Decision Altitude/Height). However, if using the ground ILS, the runway must offer a Category IIIc landing system. While Cat IIIa and Cat IIIb are common at large airports, there are no Cat IIIc runways in the world of commercial aviation.
The reason is landed aircraft wouldn't be able to safely leave the runway and taxi to the gate because taxiways are not equipped with features appropriate to no-visibility conditions.
On the other hand driverless cars now are experimented on public roads without specific equipment, including in fog, and driverless vehicles push back aircraft prior to taxiing:

Driverless pushback at Heathrow (source Youtube and manufacturer site, can be autonomous with ground markings)
In the long term airports with autonomous tugs on the whole taxi phase may be interesting as they allow savings by providing power supply while taxing, which in turn can decrease fuel needs and mechanical wear.
Is there a possibility such zero visibility landings could happen in the near future, because desirable for airlines and airports?

Confirms landing without visibility is "standard business" but that's not zero visibility. according to what I know, no aircraft is certified CAT IIIc

This has a cost for airlines, and an impact on passengers satisfaction Note that equipping all aircraft with CAT IIIc and auto-guide tractors has also a substantial impact on costs and in turn on passenger satisfaction...

@DeltaLima: This is true for many of us. I assume some customer groups would appreciate landing at the destination airport in time, regardless of the weather, the price for that wouldn't be relevant for their employers. Exactly like paying toll roads which are a profitable business (in France at least).

@mins But European Law doesn't allow airlines to make that distinction. If they are not flying because the weather prohibited it, then do not have to pay compensation to the passengers for being late. But if there were a means of taxiing in dense fog available, the weather argument no longer can be applied in that case so they become liable to compensation. Even if the costs of using that auto-guided tractor is too much for a low fare airline.

I believe Google's driverless cars rely partially on cameras, so I don't know if they could operate in zero visibility

@raptortech97: Those cars operate in uncooperative conditions. They have to read road signs, for example, and see other car's brakelights and indicators. With cooperation, they would need far less and simpler sensors. (In fact, I wouldn't be surprised if the complexity of one of those car's sensor arrays would exceed that of a modern airliner's. They typically use at least cameras in the visible spectrum, infrared cameras and LIDAR, they may also use ultrasonic sonar, radar, microwaves and passive acoustics in the audible range, plus D-GPS.)

So the pictures actually add anything, here?

Practically there are no zero visibility landings.  There is reduced visibility of varying levels.  Reduced visibility landings are only allowed when the appropriate ILS equipment is operational.
The practical issues are actually not the landing itself but separation from aircraft landing afterwards.  On the ground the runway lights are bright enough and close enough for slow navigation.  However the aircraft takes longer to clear the runway and this means increased separation between landing aircraft is required.   It is this that causes delays as aircraft waiting to land are pushed further and further back.
A ground tug to facilitate speedy removal of the aircraft from the runway makes sense in theory, but would be hazardous and the acquisition time wold be non trivial.  While an estimated stopping location could be calculated for an aircraft finishing its roll out after landing it would be subject to a very large variation and the tug would have a large distance to cover to get to the aircraft.  It would then have to connect and start the tow.  The question is whether this wold be quicker than the current arrangement and what the mass penalty would be in terms of changes to the aircraft structure to enable it to use the tug.

In response to comments I would like to clarify that my point about the acquisition time of the aircraft by the tug is concerned with the non trivial and varying distance the tug would need to cover to reach the aircraft.  Any tug configured to pull an aircraft weighing in excess of 100tons would need to be large, heavy, and have high torque to help it safely move the aircraft.  These are all things which will reduce its top speed and so increase the time it takes to get to the aircraft from some optimal starting location.

Multiple tugs:
Multiple tugs would increase the probability that the aircraft would stop within optimal distance of one of them but would also increase the collision risk and add complexity to the local airflow over the runway.
Starting locations:
To facilitate quick response and removal of the aircraft from the runway the tugs would need to be located close to hand.  However they cannot be on the runway and would be required to be back from the edge by a significant distance.  When you consider the finite number of taxiways and the sheer variation in aircraft roll outs this presents a very wide range in possible locations the tug would need to get to with a consequential increase in response times.
Aircraft Rollouts:
When an aircraft lands it typically uses a small amount of remaining velocity to keep it rolling and move off the runway.  This would almost always be quicker than waiting for a tug.
Hybrid tug system:
A hybrid system in which the aircraft uses residual speed to clear the runway and meet a tug has some advantages in terms of optimal tug positioning however it reduces the advantages in fuel consumption and increases the amount of vehicles independently moving around the airfield.

Radar and GPS should have no difficulty directing the tug to the landed plane so I don't think that's really an issue.

@DavidRicherby Radar and GPS in the aircraft should have no difficulty stopping the aircraft in one of two or three pre-defined spots with the autonomous tug nearby. Making the plane stop at a given place should be also very fast, below some 50kt the braking power is easy to determine. I think the true issue is safety: what if the plane catches fire? How are you evacuating? How do you ensure the fire and EM trucks do not kill someone? Even spraying the foam on hot brakes could be a challenge.

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text:
What is the Relation between Critical AoA with Chordline?
I read here an explanation about angle of attack (AoA) as below:
Overly long (forward-back wise) wings will generally have a smaller critical Angle of Attack while smaller ones will have more tolerance before stalling since Airflow starts to become turbulent in form of eddy currents.
I never heard that claim before neither I ever read. I have searched in this forum if someone asked that question previously, but seem is not. I could not find. Is anyone have reference to this claim?

This might help you. Similar Info

It is a bit unclear, but I think what he is trying to say is if you have longer chord, then separation of the flow will happen at lower AOA than with a shorter chord, which is plausible as the boundary layer grows thicker and thicker with longer distance. And stall=flow separation basically.

source: Stack Exchange

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sort: votes


text:
Why does Germany is not so advance in aircraft development?
I tried to see what countries are the most advance in developing aircraft. I found that Germany is not so advanced (CMMIW). Indeed, there is Dornier there beside has share in AirBus. I don't know whether they are the biggest share holder or not in AirBus.
In another side, among of the G8 countries, US is the leader. They have lot of aircraft industries. Boeing, Lockheed Martin, GulfStream, and many other. Canada has Bombardier. British has BAe with their RJ70, RJ100, so do Roll Royce with their power plant. Italy has Avanti and Augusta Westland, beside small aircraft industries. France with Dassault, Mirage, also joint venture AirBus, and some more small industries. Russia has Sukhoi, Ilyushin, Tupelov, Mikoyan, Beriev, and there are more. And Japan has Mitsubishi, Kawasaki, and Honda, now they are massively develop and  research aircraft, both for civil and military. But Germany is look like not so aggressively take part in this technology like they do in another part. Even China is now growing. With Comac, they has now C919, and now with Russia is developing C929 which will emphasizes their presence in aviation manufacture. ABC, AirBus, Boeing, and Comac are considered as the biggest manufacturer, leaving Germany behind. Is any reason why? Germans probably can explain.

Setting aside for a second the fact that, imho, this question might be off-topic. Airbus is a Franco-Spanish-German consortium with plenty of plants in Germany. You list it as one of the biggest manufacturers. Why would that leave Germany behind?

indeed, EADS has a lot of German facilities. Germany just elected to merge all or most of its aviation industry into EADS whereas France did not.

Your spelling of AirBus already implies you did no research on Airbus.

You look at big names. Germany's economy is based up on mid-sized manufacturers in general, which are know to the customers for which they matter. hph304j named already some from the aircraft industries. A lot of this mid-size companies merged already into Airbus, therefore the big name is Airbus. Looking at America, McDonnell Douglas merged into Boeing, Lockheed and Martin merged into Lockheed Martin and in Canada large parts of Bombardier are now part of - Airbus. We even create companies for single planes, Eurofighter GmbH (Typhoon) and Panavia Aircraft GmbH (Tornado).

A large part of the reason is that after WWII, Germany was prevented from having any industries that could be used to build up their military. Aircraft production was essentially banned. One of the consequences of this is that there is a robust glider industry.

I know that most of the rocket designers ended up in the US or Russia and I suspect that many of the aircraft designers did too.

From the OP's profile-- "I am nobody that one who love engines and aircrafts."

"Your spelling of AirBus already implies you did no research on Airbus"--  No, he just modified the spelling to mirror his user name

I asked this question on History.StackExchange and they gave a very detailed answer. https://history.stackexchange.com/questions/49801/what-restrictions-were-in-place-after-wwii-on-the-german-aircraft-industry/49809#49809

May i suggest, you read the wikipedia artikle "History of Airbus".History of Airbus
There are also other big aerospace companies in germany, like MTU, DIEHL and LIEBHERR just to name a few. These companies don´t build their own planes, but manufacture major components for the aerospace industrie.

Germany was in the head of Aircraft Development, aerodynamics and engines, until May 1945, when all their research was seized, industry destroyed, and constraints put in further development, not the lesser of it their envisaged market size, from taking the place F Engels described as UK goal in his 1845 work about British workers, being a manufacturer nation supplying an agricultural Europe, became impossible. If you don't have enough people buying your products, your industry can't get enough capital for R&D, this is why the population of a nation, besides commodities as energy, matters. OK?

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What is the procedure to turn an airplane when it flying?
As far as I know, there are two part devices that involved when turning an airplane when it flying: aileron and rudder. How they work? Which one is first over another when turning/banking? In this case, the airplane is the common commercial airplane which is has ailerons, horizontal and vertical stabilizers, elevators, and rudder. Not the V or /-shape empennage like mostly used in drone.

Ailerons are but one type of primary flight control surface, which are used to roll the airplane about the longitudinal axis, thereby directing the lift vector generated by the wings from the vertical to the direction of a desired turn, thereby providing the required force to cause the airplane to change direction.  Ailerons aren’t the only primary flight control surfaces which can do this; other types can include flaperons, differential elevator inputs and differential spoilers.
Rudders are used to counter adverse yaw caused by the difference in induced drag created as a byproduct of aileron inputs, which cause the nose to swing toward the outside of the turn, resulting in uncoordinated sideslip and inefficient turning of the aircraft.  In the case of a V or Swallow tail, ruddervators fulfill the function of both rudders and elevators and would be used to counter adverse yaw as well.

Do we always need to activate (to use, whatever the best word for it) rudder when turning/banking?

Not necessarily, though rudder will be needed in order to execute a COORDINATED turn.  Rudder is not needed simply to bank the aircraft, only to counter adverse yaw when rolling into and established in a coordinated turn.

Airplanes turn by banking, displacing the lift vector to an angle from vertical.  Same as a helicopter moves laterally by tilting its rotor disc, or a hovering rocket would move laterally by tilting off vertical.  The airplane is moving forward as it moves laterally, so its lateral motion is an arc instead of a straight line sideways.
The elevator's job in a turn is limited to increasing the overall lift vector, by increasing the wing's angle of attack, to compensate for the loss of the vertical lift component in the turn.  Same as a helicopter has to increase rotor thrust (or the hovering rocket its thrust) when moving laterally to keep from descending.
The rudder's job is only indirectly related to turning.  It's only job is to keep the tail lined up behind the nose in the airstream, the fin doing so passively (like a weathervane), and the rudder surface doing so actively (to counter adverse yaw from the ailerons and engine torque, which tend to overpower the weathervaning effect of the fin), while also providing the ability to move the tail out of alignment with the nose if that is desired (sideslipping).  Or to compensate for asymmetric thrust in the case of a multi-engine airplane.
If you have an airplane with zero adverse yaw effects and no torque effects, no rudder input is required at all to turn.  In a jet, there are no torque effects to speak of, and adverse yaw is very minimal, requiring very little rudder input to counteract adverse yaw.  The airplane's yaw damper system, which operates the rudder in a very limited range, has all the authority required to keep the tail lined up behind the nose when turning. You never touch the rudder pedals in a jet once airborne unless an engine quits.
Similarly, airplanes that bank by using spoilers only, like the Northrop P-61, or the Mitsubishi MU-2, and have zero adverse yaw effects, require no rudder at all in a turn.


text:
Which one is more effective produce power of this two turbine's location?
Consider we have the exactly same turbine which is possible to be mounted to the tail (left picture) and to the wing (right picture) as picture below. Which one will produce more power to the airplane? Need physics analysis.

The power will be the same. But the aircraft with wing mounted engines will have less drag. Contributing factors are:

Engines are heavy. Putting them aft means there is a big difference in centre of gravity between empty and loaded aircraft, so larger horizontal stabilizer is needed to keep it in trim over the range of centres of gravity. It is also needed for rotation on take-off with forward CoG, since the main gear must be behind the aft-most CoG position. That means more trim drag.
Putting something heavy, like engines, on the wing ahead of it reduces flutter, so the wing can be less stiff, and thus lighter. This means more payload, or less induced drag at the same payload.
Engines ahead of the wing fit nicely with the Whitcomb area rule. At the tail they add a lot of cross-section where it should be tapering off already, slightly worsening the transsonic performance.

So engines tend to be wing-mounted on all but small low-wing aircraft where they would have too little ground clearance.

The power will be more or less the same but probably favours the wing engines.  The only major difference is the pitching moment created by thrust is nose-up with underslung engines (low thrust line) and mildly nose down with the tail engines (high thrust line).
You could say the tail mounting creates a tiny bit more trim drag (because a little bit of extra tail downforce is required to counter the nose down pitching moment) than the wing mounted engines on two otherwise similar aircraft, so there is probably an argument that wing mounted engines are slightly more efficient overall because of lower trim drag losses.  The tail mounting also requires the gear to be farther aft at a less than optimum position, to keep it from tipping on its tail when empty.  This requires more tail power for rotation, with a bigger surface than would otherwise be necessary, and again slightly more drag.
So with two identical airframes, you could say the wing mounting does win out from a net power available to push the airplane along perspective, but I'm sure the difference is quite small.

More than a couple of aircraft have a tail prop-stick to prevent tipping back when empty. That could be employed to help (at least somewhat) offset the gear location.

Perhaps, but it's a really impractical solution.  If you tour a typical gate area and observe all the tail engine  airliners, Embraers, RJs, DC9s, you won't see too many tail struts.


text:
What is that extended nacelle feature on the Honda HA-420 HondaJet?

Here is a picture of the Honda HA-420 HondaJet with strange extended engine housings. Why are they required? What are they for?
I am not asking about the red covers, but the white engine housing features that I put a red circle around.

Yes, because this is the first time I saw such cover. I have never seen it in any other airplane.

Are you asking about the red GSE plugs for the jet pipes or the fairings at the end of the engine nacelles?

No, I am asking about the "cover" at 120 degree or 240 degree for another engine.

@GregHewgill, please, I am not asking the red cover. I am asking the white thing I put circle. I am not sure what is the name, is it cover or not.

They aren't covers; they are fairings to integrate the flow going past the lower part of the cowl with the exhaust stream coming out the tailpipe above and the pylon below.  It's because the cowl extends below the engine some distance to house various parts like accessories, plumbing, ducting etc.  You have to create a nice transition for the airflow or you would have a lot of unnecessary drag.
You could say that the fairing is filling in all the 3 dimensional space that the free stream and the exhaust plume can't fill in because the change in direction is too extreme, and would result in flow separation, causing turbulence/drag.

Why do they some parts only?

@AirCraftLover because having them any larger than absolutely necessary adds wetted area and, therefore, drag.

OK, tell me more. Why the position is not simply at the bottom? That position is not left/right side, but a bit shifted from buttom.

If you mean the scalloped or cupped shape of the fairing below the engine tail pipe, it's because it's accommodating the exhaust plume coming out the back of the engine.


text:
Why is the aerodynamic center for most airfoils located at the quarter chord location?
I've been reading several sources on the location of the aerodynamic center

https://web.mit.edu/16.unified/www/FALL/fluids/Lectures/f03.pdf
https://en.wikipedia.org/wiki/Aerodynamic_center
John Anderson Jr., Fundamentals of aerodynamics

Many (such as for example Wikipedia) determine the aerodynamic center by writing out the equation for the moment for an arbitrary point, and then apply some definitions to get to the end result: the aerodynamic center lies at the quarter chord, see below:

However, I'm looking for a more physically intuitive explanation, probably starting from thin airfoil theory. The way I understand it, ultimately the aerodynamic center is determined by how the distribution of lift changes with angle of attack.  Thus, the distance the center of lift moves w.r.t. the aerodynamic center cancels out the resulting moment due to the change in lift with angle in attack.
Can we explain why this point ends up to be at the quarter chord location?

It has to do with the Birnbaum distribution. The center of its area is at one quarter.
In potential flow theory, lift can be calculated as the linear superposition of a contribution from camber and one from angle of attack. While the camber-related part of lift is constant, the angle-of-attack related part varies linearly with this parameter. This means that changing the angle of attack adds or subtracts circulation, the chordwise distribution of which is described by the Birnbaum distribution. The important part is the self-similarity of the angle of attack dependent lift resulting from this addition or subtraction of the same chordwise distribution.
The center of pressure of the resulting angle-of-attack dependent part is constant and at 25% of chord for 2D flow and wings of large aspect ratio, at least as long as flow remains attached and viscous effects are negligible.
The pictures below (own work) show the circulation over chord produced by camber alone (so AoA is 0°) on the left and the Birnbaum distribution alone on the right (so using a symmetric airfoil). Total circulation is simply the sum of both. Copied over from this answer thanks to the helpful suggestion of ROIMaison.

ultimately the aerodynamic center is determined by how the distribution of lift changes with angle of attack

That is precisely right: The aerodynamic center (a better word for it is "neutral point", btw.) is the point where the additional lift force from a change in angle of attack can be summed up. Do not look at the lift distribution per se (that is what determines the center of pressure), but at its change over angle of attack, because that is what determines the neutral point and what shifts the center of pressure back and forth. The constant part caused by camber does not change with angle of attack and the part caused by angle of attack has its center always at 25% (under the conditions mentioned above) and its magnitude grows with angle of attack. This growing contribution of the angle-of-attack-dependent force is what shifts the center of pressure forward (on airfoils with positive camber) rsp. backwards (on airfoils with negative camber) towards the 25% point.
BTW, I wish we had more people like Aaron Swartz. That article is now 97 years old and these monopolists still want 42 € from me for a copy. Sorry, no details from it – I could not find any online.

I'm not sure which images you mean exactly (I do have access to the paper). However, in one of your previous answers you show the Birnbaum distribution for different angles of attack, I think it illustrates your point clearly, and would be useful to include here as well.

@ROIMaison: Thank you, I forgot about this answer completely. I did those graphs ages ago using OmniGraffle. I am a happy Inkscape user since maybe 15 years now and didn't touch OmniGraffle and those pictures for maybe 20 years.

I'm using the formulae from the Birnbaum paper to create the three parts of the circulation. If I add them, I get something that looks like the graphs you show, but if I then determine the center of pressure, it does not become the 1/4 chord point (it's just shy of 40% chord). I noticed there are coefficients in front of the three formulae, should I adjust the weights of the three functions, if yes, how?

@ROIMaison now we see how important the down-nuetral-up force of tail is to regulate CP for cambered wings as AOA increases, and also how the "plank" wing is possible on the Helios.

@ROIMaison I do not have access to the paper and only know of Birnbaum distributions from secondary literature. Sorry, but I cannot help you with those equations. And you certainly had to assure that you would not give copies of the paper away before you received it, right?

@PeterKämpf, I found an email address of you from Stabilo, is that still current?

@ROIMaison: You know my name. Now add the gmail domain and you will reach me.

@PeterKämpf I sent you an email.

@ROIMaison Did you get the Umlaut right? So far I did not receive the message.

@PeterKämpf I did not actually. I guess I made another peter happy with my email. I tried again with the new email.

@ROIMaison So far no success. What did you try? Google is quite lenient, so it does not matter if there is a dot between first and last name or if you use upper- or lowercase. Im flabbergasted that you did not get through.

@PeterKämpf
I found the issue, outlook automatically changed it back to the known wrong contact.. I resent it, let me know if it arrives this time.

@ROIMaison: Yes, success! I got it.

Are you sometimes saying center of pressure when you mean aerodynamic center? For example: "That is precisely right. Do not look at the lift distribution, but at its change over angle of attack, because that is what determines the center of pressure."

@DKNguyen The angle of attack has no influence on the aerodynamic center (which I prefer to call neutral point, btw.). Why should the sentence read differently from how it does? A change in AoA shifts only the center of pressure.

@PeterKämpf It's a bit ambiguous because it could be interpreted that how the lift distribution changes determines both center of pressure (due to change) and aerodynamic center (and due to lackthereof). That is, the fact that something changes is distinct could be considered distinct from how something changes (with the lack of change being included as one of the possible hows).

@DKNguyen: I think I get it now. Please let me know whether it is any better now.

@PeterKämpf That's a bit better. I would italicize the ADDITIONAL lift force since that is very key to the entire statement but is rather understated.

Intuitively, the aerodynamic center of a wing is at 1/4 MAC because the lift created by the front top curved surface of the wing and the front bottom of the wing roughly equals the lift created by the back bottom of the wing throughout the range of non-stalled AOA.
From this we can deduce that the top of the wing behind MAC is not a major contributor to lift.
A tail can be used as needed to help keep the net center of lift from moving too much with changes in AOA.  Tail requirement is greatly dependent on wing design.


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Is there an order for the instructions of a go-around?
Scenario: Aircraft on final is approaching the threshold. A departing aircraft is slow to roll and ATC must issue a go-around instruction.
ATC: PULL UP AND GO AROUND. TURN LEFT HEADING 360 MAINTAIN 3000'
or
ATC: PULL UP AND GO AROUND. MAINTAIN 3000' TURN LEFT HEADING 360
Is there a correct order to follow for these instructions. Set aside the reason for the go-around is obvious and turning first would appear to be prudent. In simple terms, it there anything that prevents a pilot from turning first, climbing first and then turning, or perhaps making a climbing turn?
* I'm a controller.

ATC would not tell a pilot to pull up, and would not issue any further instructions than "go around". Once the airplane on finals acknowledges the go around then ATC will issue further instructions.

Canada.  Phraseology is exactly as I stated.  If you haven't provided alternate missed approach instructions, as soon as you say pull up the aircraft starts flying the published missed.  We don't want that so, we issue a heading and an altitude.  It's the order of compliance I'm curious about.   I cannot see where guidance to pilots is written down anywhere.

@ezalpha Regardless of what is official (if anything), in practice it would be safest to be prepared for pilots to either ido them in the order stated or do both at the same time.

StephenS.  Yes, we are prepared, its just that I'm surprised that its not written down.

Logically one would expect the pilot to execute the instructions in the order given.  However, nothing precludes one from executing the instructions simultaneously based on the wording above.  If you really wanted the turn first you could/should have said:  "turn left heading 360, then climb and maintain 3000".  Then there is no doubt about the intended sequence.

Standard ICAO phraseology for ATC commanded go around for callsign XYZ would be:
"XYZ go around (possible reason for go-around)"
XYZ would then acknowledge this:
"Going around, XYZ"
ATC may then give additional instructions in the order it sees fit. If turning is first priority:
"XYZ turn left heading 360, climb and maintain 3000"
In any case the plane would of course be climbing, so if there is a heading to give, it would be logical to command it first.
If for some reason it is imperative to turn the plane away from something as quickly as possible, ATC can command "XYZ turn IMMEDIATELY left heading 360"
Why like this? Because it is imperative that that ATC can be assured as fast as possible that the aircraft will be initiating the go around. Anything other is secondary. The pilot will be preoccupied with the go around procedure for a couple of seconds, and should not worry about heading untill safe climb speed has been established. It is also good practise to keep messages short when something out of the ordinary is happening. Not that go arounds would be that rare, but the pilot might be wet behind the ears, and too many instructions might overwhelm him/her.
In any case ATC must assume there will be some delay in anything an airliner does. There's all kinds of CRM going on in the cockpit to verify everything happens safely.
As to why Canadian phraseology would use the non-standard "pull up" seems strange, but from what I hear Canadians are very nice people so I'll let it slide :)

Set aside the "pull up" vs "go around" part.  This happened. 787 3/4 mile final had to be pulled because a 737 was slow to roll.  787 was told to maintain 3000, turn left (50 degrees).  Controller expected the aircraft to turn left almost immediately (it had obstacle clx) and start climbing.  Instead, it climbed straight ahead.  Amazing how quickly an empty 787 accelerates.  It darn near passed the 737 before we got it sorted out.  Crew report says F/O (flying) started to turn but Capt said "no we climb to 3000 then turn".  Capt won.  My question is: who's right?  I can't find an answer

What airfield, rwy @ezalpha? Maybe the published missed approach has something about the climb part. If instructions were as you stated, I think it is not unreasonable from the capt. to think you expect to first climb to the altitude, then turn. If you want immediate turn, you command an immediate turn.

Parallel ops, but that's not the point (ie divergence).  We said turn left maintain 3000.  A/c was 1100 agl, obstacle clearance not an issue.  Issued at 3/4 mile we expect the aircraft to be turning before it reached the threshold.  Instead, it overflew most of the runway. FO (pf) started to turn.  Capt overruled.  They discussed on the ground.  There was another pilot in the jumpseat.  He agreed with the first officer.  To my point.  We didn't say "as soon as able" or "immediately".  I think they can turn, climb or a combination.  Shouldn't be confusion over something as basic as this.

Cockpit is not a parliament @ezalpha, there's no voting when PIC makes a decision. I bet the reason to climb straight was because of a published missed approach. As I don't know where exactly this happend, can't be sure.


text:
Trying to identify large WWII aircraft by aileron position on rudder
So I have this artifact, a man's slide-on tie clip with the figure of a large, overwing, possibly four engine aircraft, with possible pontoons. Any idea what this baby is? It looks like a Pan Am Clipper, but the position of the aileron, midway up the rudder, has my search skills stumped so far. Thanks.

Welcome to aviation.SE! Can you post a picture of the clip?

The aileron is midway up the rudder? Do you mean the horizontal stabilizer (or tailplane) is midway up the vertical stabilizer (or tail fin)? Ailerons are way out by the wingtips; they're nowhere near the tail, where the rudder is.

Please try to fix the question to address the incorrect use of "aileron".  A photo of the artifact would also be extremely helpful.

The Boeing 314 Clipper had a very distinctive triple tail, but it also had (what appears to be) a trim tab half way up the rudders, so maybe that's it?

The PBY-5A has retractable wing tip “pontoons”, and meets all you criteria except “possible four engine”.
It is possible that the tie clip is just a generic representation of a flying boat, and may not represent an actual aircraft.

Spruce Goose, maybe?

Wikipedia: Hughes H-4 Hercules
It's not exactly WWII, but close, first (and only) flight was in 1947.
P.S. I guess by aileron you mean elevator ;)


text:
What is the selection algorithm in f-16 stick LVDT?
In a reference:
https://www.nasa.gov/centers/dryden/pdf/88102main_H-1344.pdf
page 67 (71 of pdf file) in Table 12,
it is said that the signal selection algorithm from the redundant LVDT of stick in f-16 is GCAF which means good channel average plus monitoring for a fourth input. What does this sentence exactly mean? I know the meaning of good channel average but what is the meaning of good channel average plus monitoring for a fourth input?


text:
How should the maximum output force of the actuator be chosen?
If the hinge moments of the control surfaces of an aircraft are given, how should the maximum output force of the actuator be chosen? Is there any standard or reference?

Wouldn't this depend on: a) how fast you want the surfaces to move and b) how much weight you can afford for the actuators and their fittings, which will be heavier the higher the forces involved?

If the largest hinge moment is known for the entire flight envelope, including that required for continued safe flight and landing after failure(s), then the actuator can be readily sized. Of course, some engineering judgement is needed to compromise between:

Buffering for additional unforeseen loading (i.e. factor of safety) for later stages of the design and flight test, during which swapping the actuators can be costly, and
Sizing down the actuators to save weight and costs.

Therefore, the more scenarios you have considered for obtaining this maximum hinge moment and the more confidence you have in your model, the better off you are. For example, for elevator, consider the following (but not limited to):

Achieve -1g to 2.5g at Vmo/Mmo with 3 sec out-of-trim in the opposite direction (14 CFR 25.255(a) and (c))
Achieve 1.5g at Vdf/Mdf with 3 sec out-of-trim in the opposite direction (14 CFR 25.255(a) and (f))
Takeoff rotation with dual hydraulic failure
Achieve 0.8g and 1.3g at Vmo/Mmo with dual hydraulic failure (or single elevator surface jam)

Thanks for the response. Do you know a reference that i can study for this purpose?

@shooschool Are you referring to the hinge moment requirements?

Where do the "3sec" comes from? The g load given seems to come from maximum acceptable for airliners, but i think you should develop why this values.

I have the hinge moments. I need to select actuator for them. I want to know how much force should the actuator be able to provide.

@ManuH Edited my response to include the sources.

@shooshool Moment = ram force*arm, no?

On the CRJ200, it originally went into test with the 2-actuator elevator system of the Challenger from which it is derived, and it was found the surface was too small, so some chord was added, then it was discovered the 2 PCUs were lagging, giving spongy elevator feel.  So it was decided to add a 3rd PCU to each elevator, tapping off the 3 hyd system, in a panic 6 month redesign.  On the other hand, it was found the rudder's 3 PCUs would stall at partial travel above a certain speed, but this was found to be a plus because it eliminated the need for a rudder travel limiter and was left as-is.

@JohnK 3 PCUs per elevator? Yikes.

Lots of them are like that.  All the RJ line.  Also 767, which is at a very similar overall technology level systems wise, being products of the 70s/80s.


text:
What hardware standards should the flight control computer pass?
There are some standards for each part of the aircraft like actuators, sensors,...
I could not find one for the flight control computer hardware. For example what are the environmental tests, or any standards for its connectors and ports,...?

Are you looking for design & product validation or installation/system validation?

I mean design & product validation

Based on your comments on @Gerry's answer, are you looking to develop for civilian or military aircraft?

In the US for commercial aircraft, SAE-ITC provides recommended design standards in ARINC Characteristic 701-1 Flight Control Computer System (FCCS).
The recommended environmental standards are in Attachment 6 of ARINC 701. This references environmental categories defined in RTCA DO-160. The issue is that ARINC 701 references an earlier version of DO-160 than the current DO-160G.
The main reason ARINC 701 has not been kept current (701-1 was the last update in 1983) is that the aircraft manufacturers consider the FCC an integral part of the aircraft and they provide the detail design specifications to their FCC supplier. The specifications typically reference DO-160 (whatever version is current when designing the aircraft). The actual categories and test levels are specified based on their analysis of the environment where the equipment is installed in the aircraft.
To summarize; the FCC hardware will have to meet the customer's specification which is usually in terms of RTCA DO-160 tests and categories. The actual levels are unique to each aircraft.

How about the military aircrafts? Do they use DO-160 too?

The military has been slowly migrating away from MIL standards towards industry standards so there are times when DO-160 is specified. Most of the time they do use the traditional MIL-STD-810 Environmental Test and Engineering Processes and MIL-STD-461 Requirements for the Control Of Electromagnetic Interference Characteristics of Subsystems and Equipment. The biggest difference between them is that MIL-STD-461 includes additional requirements for external EMI threats that are relevant only to combat environments.

If I want to obtain requirements for flight control computer from MIL-STD-461 which part of it should I use? Does all parts apply?

You need to look up MIL-PRF-38535J which specifies electronics performance requirements. Toward the late 90s/early 00s the "performance standard" or PRF, replaced MIL-SPEC for the manufacture of ICs and microprocessors, and production of components built to MIL-SPEC (which specifies physical construction requirements) was discontinued.
MIL-PRF doesn't require specific manufacturing specifications, only that components can meet a performance standard.  This allows manufacturers to use commercial parts as long as they meet the standard, and they will find compliant components by performance testing individual parts from a batch in accordance with MIL-PRF, and set aside those that pass for use.
At the OEM I worked, our engineering organization was notified by suppliers in the early 00s that they were switching to commercial ICs/Mircroprocessors filtered by MIL-PRF testing for manufacture of various cards in black boxes because MIL-SPEC hardware was being discontinued, and if we didn't like it too-bad-so-sad there wasn't really any choice in the matter.
For individual bits like connectors and ports, a supplier will use whatever commercial parts suits their fancy, and the validation is done with when the certification testing (environmental, endurance etc.) of the black box is performed to satisfy the OEM's agreed-upon requirements.


text:
Does the SAS (stability augmentation system) actuators of f-14 use jet-pipe or flapper-nozzle technology?
F-14 uses electrohydraulic servo actuators for its SAS. What type of hydraulic amplification is used in F-14? jet-pipe, flapper-nozzle or something else?


text:
what are the no load rates of f-16 control surfaces?
What are the no-load rates for the control surfaces of f-16 like flaperon,leading edge flap and horizontal and vertical stabs?

You may expand the meaning of "no-load rates"

@ymb1 is that the definition of "no loads rate"? if so, it should be included in the question's description


text:
Does Smeaton's coefficient have a modern accepted value or it is dependent of the air density?

Does Smeaton's coefficient, k, have a modern value or it is
dependent of the air density?
Why is the accepted value of k so high?

In various texts about the Wright brothers (see 1 and 2) one can read about Smeaton's coefficient that troubled them a lot and that they finally discovered the parameter had a much lower value reaching the conclusion $k = 0.0033 lbf/ft^2/mph^2 = 0.79 kg/m^3$ (instead of $k = 0.005$), a fact also noticed by others before them.

"the Wright brothers calculated a new average value of 0.0033.
Modern aerodynamicists have confirmed this figure to be accurate
within a few percent." Source: Correcting Smeaton's
Coefficient
$L = k \cdot  S \cdot V^2 \cdot C_L$
$L$ = lift in pounds
$k$ = coefficient of air pressure (Smeaton coefficient)
$S$ = total area of lifting surface in square feet
$V$ = velocity (headwind plus ground speed) in miles per hour
$C_L$ = coefficient of lift (varies with wing shape)"

Source: The Wiki page of the Wright brothers
However, knowing that the modern formula for lift is
$$L = 0.5 \cdot \rho \cdot S \cdot V^2 \cdot C_L$$
Where $\rho$ = the air density.
It appears that $k = 0.5 \cdot \rho$ and so it does not have a standard average value. Also a $k = 0.0033 lbf/ft^2/mph^2 = 0.79 kg/m^3$ leads to a $\rho = k/0.5 = 1.58 kg/m^3$ that corresponds to a sea level air temperature well below -25 C, which is unusual.
If the two relations for lift are correct, the Smeaton's coefficient can not be 0.0033 but closer to 0.0025 a value corresponding to a standard air density at $20 ^\circ C$ close to $1.2 kg/m^3$.

Smeaton's coefficient is only valid in Middle Earth.

@Koyovis I believe you're referring to Smeagle's (or Smeagol's) coefficient - which begs the question from Annie in The Martian (in the book but not the movie) about Project Elrond... but that's a whole different discussion!

Wright brothers used the following 1900s equation:
$$
L = k \cdot V^2 \cdot S \cdot C_L
$$
$k$ = pressure force (drag) on a one foot square flat plate moving at one mile per hour through the air.
$$
D = k \cdot V^2 \cdot S \cdot C_D \tag{1}
$$
Modern drag equation is as follows:
$$
D = \frac{1}{2} \cdot \rho \cdot V^2 \cdot S \cdot C_D \tag{2}
$$
Therefore, by equating $(1)$ and $(2)$, we get:
\begin{align} k   &= \frac{1}{2} \cdot \rho \\\\ & = \frac{1}{2} \cdot 1.225 \\\\ & = 0.6125 \left[\frac{kg}{m^3}\right] \end{align}
Value of k was given in units of lb force/m.p.hr sq/ft sq
1 lb force/m.p.hr sq /ft sq = 0.45359237 Kg × 9.8066 m/sec sq /(0.44704 m/sec) sq / 0.0929 m sq
=239.5939 kg/m3
Therefore
k = 0.6125/239.5939 = 0.00256 lbf/m.p.hr sq./ft sq.
Thus modern value of k is 0.00256 if we take air density as 1.225 kg/m3

There is no modern equivalent to the Smeaton coefficient, therefore it has no "modern" value.
The Smeaton coefficient is supposed to represent a "coefficient of air pressure" (and thus indirectly density), but it is a false constant: There is no single value because air pressure is not a constant value, but rather a variable (depending on altitude, temperature, etc.).
As you noted lift is calculated by a different formula today, using actual density of the fluid (air) you're moving through in place of the Smeaton coefficient.
$ L = Cl \cdot \frac{(A \cdot \rho)}{2} \cdot V^2$
You can calculate the Smeaton coefficient at any given location/altitude with simple algebra if you know the air density: Solve the modern lift equation, plug the values into the "old" Wright Brothers era equation, and solve for $k$.
The density-based calculation also has the benefit of working in other fluids besides air. (The same equation can give you hydrodynamic lift for boats and submarines by plugging in the density of water - itself a dynamic value depending on temperature, salinity, etc.)

If the formula L = k x S x V^2 x CL is correct (I suspect it is not) then the Smeaton coefficient can not be 0.0033 unless the Wright brothers experimented in temperatures well below -25 C.

@Energizer777 the formula is correct, my understanding is the coefficient is based on drag using a flat plate & their measurement was pretty close to what you'd get if you worked the math, just poorly compensated (temperature & altitude/pressure change the "constant")

And it's also useful if the air is, say, cytherean or jovian or solar air rather than terran air!

Your grouping of the terms in the equation is a bit unusual. The terms for dynamic pressure, $\frac12\rho V^2$, and the dimensional coefficient of lift, $C_l A$. are usually kept separate. Note that the $\frac12$ comes from the definition of kinetic energy, which is divided by volume to get the dynamic pressure, so it makes sense to apply it to the velocity, not the area.

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/


text:
What are the definitions of elevator angle of attack and critical angle of attack?
Angle of attack is the angle between the chord line of the wing to the direction of relative wind. So angle of attack helps the aircraft climb.

Then what is the angle between the chord line of the elevator and relative wind? Because if I am not wrong, by pitching the elevator up, the wing of an aircraft is subjected to the 'angle of attack'.
When you say angle of attack exceeds the critical angle of attack, is it because the elevator is pitched up thus creating wider angle between the chord line and the relative direction of the wind?
In the Air France 447 accident, the PF kept saying, keeps the wings level. What does it mean? How can the wings be leveled when the angle of attack is so high? Does keeping the wings level mean reducing the angle of attack to zero?

relevant question about the critical AoA: https://aviation.stackexchange.com/q/6366/1467

angle of attach is not helping the aircraft climb. Airplane can be climbing with negative angle of attack. What makes the airplane climb is the angle between the airplane speed and the ground, called climb rate. Then the speed might have another angle relative to the airplane main chord

@Trebia Project: didn't quite understand what did you mean by angle between airplane speed and ground.

He means that if the aircraft velocity points away from the ground, the aircraft climbs, even if the AoA is negative in that moment.

be very carefully with "angle of attack helps the aircraft climb". It is a very common misconception and it takes ages to convince someone of the contrary: In a climb the lift (and thus load factor and angle of attack) is less than in level flight. Google for lift force in a climb and see why.

Angle of Attack is the angle between the chord line of the airfoil and the relative wind. The elevator is also a "wing" in the effect that it is an airfoil.

1) Then what is the angle between the chord line of the elevator and relative wind? Because if I am not wrong, by pitching the elevator up, the wing of an aircraft is subjected to the 'angle of attack'.

It's also called "Angle Of Attack". Pitching the elevator up also changes the angle of attack of the elevator. There are such conditions as elevator stall, although much less common than wing stall.

2) When you say angle of attack exceed critical angle of attack, is it because the elevator is pitched up thus creating wider angle between the chord line and the relative direction of the wind?

The "Critical AoA" is the angle at which the airfoil is generating the most lift.

[Wikipedia Article on AoA]: The critical angle of attack is the angle of attack which produces maximum lift coefficient. This is also called the "stall angle of attack". Below the critical angle of attack, as the angle of attack increases, the coefficient of lift (Cl) increases. At the same time, above the critical angle of attack, as angle of attack increases, the air begins to flow less smoothly over the upper surface of the airfoil and begins to separate from the upper surface

Usually when you exceed the critical AoA, the air on the upper surface of the airfoil de-laminates and becomes turbulent. This results in a loss of lift (stall).

Source: FlightSimBooks.com

3)In the Air France 447 accident, the PF kept saying, keeps the wings level. What does it mean? How can the wings be leveled when the angle of attack is so high? Does keeping the wings level means reducing the angle of attack to zero?

I'm not sure what context you are talking about. I've read the transcripts pretty carefully and I don't recall the PF saying to keep the wings level. The Captain said to descend, while the PF was trying to climb. The SIC was pushing the stick forward while the PF was pulling back, causing the wings to remain level (or at a slight climb).
Keeping the wings level (avoiding an un-coordinated turn) in a stall condition is the best way to avoid putting the aircraft into a spin. This is possibly what they meant, although they didn't know that they were in a stall (or if they did, they were not reacting to it appropriately).
Yes, leveling the wings would be making the AoA at zero (zero meaning where the airfoil is developing lift, and depending on the bank angle), but only if there were positive airspeed to make the relative wind parallel with the chord. In the case of AF447, the aircraft was falling out of the sky at a spectacular 15,000 FPM, which means that the relative wind to the airfoil was basically perpendicular. They didn't have the forward airspeed to change the AoA, so they should have lowered the nose to gain airspeed and decrease the AoA.
The important thing to remember is that the pitch angle of the aircraft is not the same as the AoA. You can take an aircraft at zero forward speed and drop it straight down, the pitch angle will be zero but the AoA will be perpendicular.

I believe that keep the wings level does not refer to pitch but to roll.

@Simon Yes, I edited that in. AF447 hit the water at about 10° of pitch, but they did have a lot of issues with roll control early on when the system switched control laws. I'm not sure where in the transcripts the OP is referring to.

I've just searched the transcript: there is a mention of wings level @ 2 h 12 min 20,8 and from 2 h 12 min 54,5 till 2 h 13 min 05,9


text:
What are these parts on the wing of a Boeing 737?

In the picture above:
What are the parts numbered "1" and and "3"?
Is "2" maybe the reinforcement of the pylon?

In the background: London Gatwick Airport's reasonably new skybridge. Fun to walk over and watch airliners taxi beneath you.

I try to improve on the guesswork of @aeroalias. Here we go:
Parts 1 and 3 are the two exhaust gates on either side of the aft engine fairing, which is part number 2. For a reference, see The Boeing 737 Technical Guide, page 116, by Chris Brady (Google Books link).
The flap actuation mechanism is quite a bit ahead of the flaps, and the same goes for the spoilers.
The engine on the Boeing 737 is placed very close to the wing, so the flaps would sink into the jet blast when lowered to the landing position. This would cause high vibrational and thermal loads on the flaps next to the engine exhaust, especially when the pilot increases thrust in landing configuration for a go-around. To protect the flaps, the exhaust gates were added so this part of the flap does not extend down as much as the rest of the wing flaps.

Drawing of the inboard wing flap of the Boeing 737-300/400/500. The inboard flap track sits within the wing root fairing, and the outer track sits in the aft engine fairing (picture source).

Numbers 1 and 3 are exhaust gates to prevent damage to flaps due to engine exhaust (thanks to Peter Kämpf). Here they are in deployed condition.

Source: britmodeller.com
Number 2 is the flap track fairing. it contains the flap operating mechanism.
The same thing, from bottom:

Source: www.b737.org.uk

But in this position there is no flap.

The flap is most certainly there, it's just not extended. You can see the flap track within the fairing. The fairing (closest to the plane's body) is white, and has a silver arc cutting through it - that's the flap track.

Well, there's not a flap where the track fairing is, but there are flaps adjacent to it on both sides. The first picture in this answer shows well what it looks like when it's extended. Also, it looks like 1 and 3 are the parts of the flap adjacent to the track fairing. The spoilers are farther forward.

Would you PLEASE refrain from posting your guesswork when evidence is easy to find online? You were wrong again, -1.

@PeterKämpf Looks like the cutaway I referred was labeled wrongly.  Thanks for pointing it out. Corrected the answer.


text:
What is the function of activating an individual elevator on the B-1B Lancer?

This B-1B Lancer's picture I captured from this Youtube video. The control axes is from Wikipedia.
As we know, the three "conventional" axes are as follows:

Roll axis controlled by aileron,
Pitch axis controlled by elevators (both at the same time),
Yaw axis controlled by rudder.

Then my questions are:

What is exactly that purpose (the single elevator activation), and
How it work? Is there new axis of it?

That would be the correct control response for a combined roll to the right and nose up pitch command from the pilot. It looks like that photograph was taken while the pilot was in the middle of doing a check of their control services prior to taxi.
You can see similar results in the tower buzz sequence out of Top Gun. The final bit shows an F-14 from the tail which rolls to the right and pitches nose up with a similar control response from a combination of the tailplane and starboard wing spoilers.

Is it more efficient to use differential elevator deflection for roll (in addition to the normal use of elevators and ailerons) than to use the elevators only for pitch and the ailerons only for roll?

@Tanner Swett It depends on the speed and flight profile.  Ailerons offer good roll authority in Low subsonic flight while differential spoilers and all moving tailplane work better in high subsonic and supersonic flight regimes.  I’d say differential spoilers offer the best roll authority but combinations of control surfaces are frequently used to increase maneuverability and maximize control authority.

@ymb1 I prefer the term all moving tailplane in these kinds of applications because it offers both control in roll and pitch.  Stabilator is also common but there is a considerable difference in operation between a PA-28 or C-177 stabilator and that found on say an F-16.

@Tanner - ailerons have move leverage than elevons, but fighters generally have fairly short wings so the difference is small, and the designers wanted to use the whole trailing edge of the wing for flaps.

Upvote for citing Top Gun

@CarloFelicione, yes it is. That jet is in preparation to fly. But actually I assume it during flight, what is actually that function of activating single elevator and how it works? Quite different from the Hornet as asked in your link.

@CarloFelicione, I believe all control must be come from pilot. So, consider the jet is being fly, where the jet will point by activating such elevator? Which axis will be affected. I assume the pitch (up) and the roll (to the right) axises will be affected. I am not sure about yaw axis.


text:
What is lifting force of V-22 Osprey? Is any other than the two rotors?

Picture is from here.
Is any other force to lift the V22 other than its rotors? Is any possibility that the power plants their self are not generating thrust? I was just wondered that such short propellers can lift (especially when VTOL) the aircraft even with additional heavy humvee.

In horizontal flight, the wings generate all of the lifting force while the rotors are just used for thrust.

The lift is provided by the rotors only. The rotors are powered by Rolls-Royce T406 engines, which are turboshafts: all useful generated turbine power is converted into shaft power for the rotors. Note that turboprops do convert some of the exhaust energy directly into propulsion thrust, turboshafts do not.
All helicopters can provide enough rotor thrust to lift their own weight plus that of (internal or external) payload. Wikipedia lists the max. VTOL weight as 23,859 kg, about the same as the CH-47 Chinook double rotor helicopter at 22,680 kg. The Chinook rotors are indeed longer and therefore require less power to generate the same amount of thrust: 4,590 kW for the V-22, 3,529 kW for the Chinook.

The Chinook rotors are indeed longer and therefore require less power to generate the same amount of thrust: 4,590 kW for the V-22, 3,529 kW for the Chinook, is that mean that the V-22's is rotate faster? Do you know what is the rotor speed (RPM) for the V-22 when VTOL with the 23,859 kg load?

Excellent illustration of lift gain in efficiency from higher aspect Chinook rotors! (over 25%!).  The Osprey's shorter rotors are more suitable for faster forward flight.

All turbine power is not converted to shaft power. The efficiency is not 100%, as all of the kinetic energy of the exhaust gas is not converted to shaft power. Some escapes and creates (neglible, as I stated in my answer) thrust.

@Jpe61 Your answer confuses turboprops with turboshafts. The exhaust heat causes problems for ships decks. A combustion engine obviously has an exhaust, the question in the OP was: does any of it contribute to the useful thrust, which is not the case.

@AirCraftLover The defining factor is the rotor tip speed, which needs to remain below the drag divergence speed. A rotor with smaller blades can and does turn faster. At a tip speed of M0.9 and a blade radius of 5.8m, rotation speed would be around 500 RPM.

@Koyovis it makes no difference, turboprop or turboshaft, there will be residual thrust from the turbine. The original questions asks whether or not the engines generate thrust. They do, but, once again, very, very little.

@Jpe61 Please inform yourself a bit better on the differences between turboshaft and turboprop. We're talking about an aircraft here. The pilot also creates thrust when he whistles a tune, it won't affect the payload capacity very much though.

Feel free to enlighten us here https://aviation.stackexchange.com/questions/16177/what-are-the-differences-between-a-helicopter-engine-turboshaft-and-an-aircraf @Koyovis.

The engines do create some thrust (exhaust gasses), but it is of little significance. Most of the energy from the turbine engine is harvested to turn the propeller as explained on NASA website (for example):
NASA: Turboprop Thrust
The exhaust of Osprey engine has  actually proven to be quite problematic, as it damages the take-off platform:
Wikipedia: Bell Boeing V-22 Osprey (see design section, last paragraph)

What is the meaning of this phrase Due to the requirement for folding rotors, their 38-foot diameter is 5 feet less than optimal for vertical takeoff, resulting in high disk loading, especially the phrase 5 feet less than optimal for vertical takeoff?

Longer blades are more efficient (this would make a great question itself), the Osprey cannot use any longer ones because the wing-engine-rotor combo has to be able to fold in for stowage during carrier operations. Picture here: https://images.app.goo.gl/oa2wzJkW26S1yKcH6

I think, it is not the case. They use the current size was due to proper calculation. They could extended the wing if it is necessary. This video shows  perfectly how it folded. https://www.youtube.com/watch?v=_45aUrES-j0. There is a bit more space that can be extended without extended the wing.

Of course longer blade will be more efficient. But, the question is, why they used such short blades if (as I commented above) they still can extend? Of course there was reasons.

On your first reply you said yourself they did proper calculation. If they did proper calculations, I bet they used the rotors the calculations told them to use. They were optimizing two things: VTOL performance and cruise performance. The result is what we see on V-22. And: 5 feet longer rotor blades = 5 feet longer wing = tail moves 5 feet backwards and wing takes 5 feet more room in the front as well. You end up with 10 feet longer aircraft. Maybe thats just a tad too long, I don't know.

As in all turboprops, some thrust is derived from the turbine exhaust, but it's very little, compared with the big thrust of those enormous propellers...


text:
Is any device installed on airplane to measure wind speed relative to the ground, and its direction?

A device or tool which can measure actual wind speed to the ground rather than to the airplane. Also a device which detect the wind direction to the plane. The device I meant is something like this pitot. But unfortunately this pitot is measures the airplane relative to the air around it, not the ground.

The "wind speed relative to the ground" would just be the (vector) difference between ground speed and true air speed:
 (image source)
Most aircraft do not measure ground speed and track directly with instruments installed on the aircraft. It can be measured with a Doppler radar (see Peter Smith's answer), which is only used on military aircraft (mostly helicopters) today.
Ground speed and track can be determined with external tools:

DME: The Distance Measuring Equipment can be used to measure distances to fixed radio beacons equipped with DME (usually VOR/DME or VORTAC). If you know the angle at which you are flying towards/away from/to this station, you can calculate ground speed from the change in distance. Some DME instruments display this:

(image source)
When tracking a VOR, you also know the true track angle required for computing the wind direction.
GNSS: A Global Navigation Satellite System like e.g. GPS allows precise determination of the aircraft's position. The ground speed can be calculated from the change of this position and the true track is simply the direction of change. A more accurate determination is possible by measuring the Doppler shift of the signal (thanks to bogl for pointing this out).
ADIRS: An Air Data Inertial Reference System as installed on airliners knows the aircraft's attitude and position based on integrating data from accelerometers and gyroscopes. It can be updated via GNSS, but would still provide data for ground speed and track without it.

Modern avionics can display the result of these computations directly:

left: Boeing 737 Navigation display (source), right: G1000 (source)

There are a lot of instruments that calculate the wind speed/direction. For example it is available on the G1000 (you can see it in this picture, the box with the arrow and number).

@RonBeyer I assume the G1000 would require a GPS signal to do this calculation?

It is a "sensor fusion" type device, taking into account the heading, ground track, air speed and ground speed. Pilots are taught how to do this with an E6B and using other means than just GPS, you could do it just by dead reckoning and a compass if you wanted to. I'm sure the G1000 uses the GPS though.

@RonBeyer thanks, I added the image as an example.

The wind direction and speed relative to the ground can also be calculated if the aircraft is fitted with a doppler navigation system. This measures the actual track of the aircraft and is common in military helicopters and some older military fixed wing aircraft.

Most Soviet airliners (including those still operating) had a Doppler radar.

The cheap way is by means of a drift meter, which is a telescope that points straight down towards the ground and which contains a set of crosshairs that can be rotated. The navigator looks through the telescope and rotates the crosshairs so as to determine the angle between the plane's movement through the air and its movement over the ground. From this she can calculate the wind triangle. This is how it was done before doppler radar and other electronic navaids were invented.

A Doppler navigation system based on the doppler radar principle provides the actual ground track of the aircraft.
Such systems require at least 3 beams and often are 4 beam (the 4 beam system fitted on the Wessex 1 was a high precision unit used in seaborne rescue - overcoming the backscatter from the water and waves makes for a challenging design).
The Sea King HAS 1 had a 3 beam system which actually fed ground track information to the ARI5955 Radar display (cannot find a suitable link) so the operator could view a moving aircraft across a static terrain map (very interesting and new for the time - early 70s).
The basic principle is that the beams are transmitted at 45 degrees, -45 degrees and -135 degrees (and +135 degrees for a 4 beam solution) to form a cross:

The beams are aimed downward so they measure the doppler effect relative to the ground; each beam will have sine and cosine elements; if they are equal for all the beams then there is no drift; of they are not equal, the actual sine and cosine outputs can be effectively added to find the actual ground track.
From there, if we know:

Airspeed
Ground track
Ground speed

We can compute the aircraft drift angle and therefore the wind speed and direction relative to the ground.
Although the 4 beam system is rather intuitive, we actually only need 3 beams to compute the drift angle.
Dopplers do have susceptibilities (much of which depends on the downward beam angle), but the advantage of being totally self-contained is quite large.

As far as I know, airplanes just use external data for winds aloft. GPS allows to measure ground speed and true course over the ground, that's good enough.
The closest thing to measuring actual wind speed at altitude is GPS drift in a balloon, or gliding flight computer that detects circling flight (using GPS) and calculates drift.

This isn't correct. The Flight Management Computer (FMC) in an aircraft will take the heading and true airspeed and compare that vector to the actual track over the ground from the INS; the difference in those two vectors is the effect of the wind.  Aircraft use external data -- i.e. forecasts -- when predicting ETA's and such to waypoints ahead, but the winds "now" are calculated internally.

Fair enough, I'm talking about simple GA planes. Wonder if an airliner heading sensor is precise enough to not introduce substantial error...

The heading from an INS unit is extremely accurate; the winds out of an FMC calculation are substantially more accurate than most forecasts.


text:
What is the explanation of this airfoil phenomenon? Why the air from below the leading edge's tip deflected up?

I have difficulties to understand this air phenomenon, why the air from below the leading edge's tip (from below the horizontal line) is deflected up and not just follow the bottom wing's surface? What is the physics explanation of it?

You may understand air molecules travel in all directions and what you see is the trajectory of small particules. This trajectory is the result of air particule hitting this particule, i.e. an average o  a small volume if air around this particule.

The short answer is: because not all of the incoming air can fit under the airfoil, so some of it gets pushed up and above the leading edge. This air could also fit under the airfoil by other means like becoming denser (compressing), but that requires more energy, so it mostly flows around.
In your image, the effect is exaggerated by the closeness of the ground, which places the entire airfoil in ground effect. In normal flight conditions, without the ground blocking it, more air would be able to flow underneath the airfoil.

The longer answer has to do with the speed of sound, which is the speed at which pressure waves propagate through a fluid. If the free stream is subsonic, pressure waves generated by an object can flow upstream and alter the flow ahead of any disturbance.
This is exactly what is happening here: the airfoil is affecting the pressure field, specifically by creating an area of high pressure underneath. This affects the incoming flow, deflecting it upwards ahead of the airfoil.
If this experiment were performed in supersonic conditions, you would not see any disturbance in the incoming airflow until it hit the bow shock wave of the object, after which it would become locally subsonic and be abruptly deflected to flow around the airfoil.

"Pressure waves propagating through a fluid"?  Perhaps, subsonicly, a mass of higher pressure (therefor denser) air forms under the pitched wing, which deflects incoming air. And let's not forget the lower pressure behind the wing, which also helps deflect the flow, and accelerate it, over the top of the wing (see complete smoke trail).

@RobertDiGiovanni: At subsonic speeds variations in density are small. High pressure mostly means slow moving, not dense. Big changes in density are only possible in supersonic flow.

@Peter Kampf its not an all or nothing rule, only greatly magnified at higher speeds (who can really fathom 550 mph?).  Of interest is how the higher pressure (ok, not density) and lower pressure affects the air flow more than the form itself, although the form creates the highs and lows.

The picture is from here, a video of spoiler of a car's testing. It is spoiler tunnel test with smoke. Hard to say that the below of the wing is "congested" or compressed.

@AirCraftLover hard to say, because car spoilers are mostly fashion accessories outside of high-performance competitions like F1. The tests in that video are at completely wrong Reynolds numbers, but they already show the main issue: the spoilers are immersed in detached flow from the car body, and thus very inefficient. But that is essentially a whole different question.

@AirCraftLover To answer you, imagine if you extended the spoiler in your picture by a few cm until it touched the ground; then you'd have no questions about why the air is going over it, because it would have nowhere else to go. The space underneath is a bottleneck, so the flow moves through the easier path above the airfoil.

If you check the video in 5:44, it is clearly shown that there was tendency the smoke deflected upward. I did intentionally to capture the extreme AoA to show the deflection upward tendency. But with shallow AoA as shown in 5:44, the smoke also deflected up.

@AirCraftLover at 5:44 in the video the smoke moves up because the airfoil is in the way. As Peter Kämpf writes in his answer, there exists an imaginary plane, the air above which will go over the airfoil. It just happens that this plane is lower than you imagined, but I have already given you the why: the presence of the airfoil creates zones of different pressure, and the airflow reacts to them, moving away from high pressures and towards low ones.

@AirCraftLover Unless your question is why does the air go up when I was told airfoils push it down? In that case, this is a case of local versus global effects: it goes up first, but then goes down more, downstream of the airfoil, only this is not clearly shown in the video (there are multiple problems with the experimental setup in that video, don't look too deep into it).

Simple answer: The plane that divides the air that goes below the airfoil from the air that goes above sits quite a bit below the leading edge.
There are several things to see here:

Local speed: The closer and more focused the smoke lines are, the higher the local flow speed.
Local pressure: Speed equals lower pressure, so the area of the closely packed smoke lines also indicates very low pressure. Bernoulli's law applies here (along one stream line).

This low pressure is sucking in air from all around, and that is what bends the smoke lines ahead of the airfoil up. Of course, pressure always means a pressure difference and this effect is magnified by the blocking of the flow near the trailing edge. As @AEhere correctly observes, ground effect creates a high pressure area ahead of the lower wing and along its length which pushes the smoke lines up. At the gap between trailing edge and ground the flow speeds up again, but unfortunately there are no smoke lines to illustrate this.
The diffuse smoke lines past the nose region show that the flow slows down there and turbulence mixes the smoke with more air. The flow separates at the nose where a wedge of stagnant air helps the smoke lines to move away from the airfoil contour. Pressure from above pushes the diffuse smoke down over the rear airfoil so the smoke past the trailing edge sits on top of the flow which manages to squeeze through between the trailing edge and the ground.
Without the ground effect the upwards bending of the smoke lines would look quite similar but more air from below would fill up the area behind the airfoil so the smoke would not move quite so much down.

Condensed water vapor from the nacelle vortex generator negotiating the leading edge of a Boeing 737 (picture source). Here the nacelle intensifies the effect, but the direction of the extended slats indicates that the local flow direction along the wing span does not differ that much from the flow visualized by the condensed water vapor.
Here the flow stays attached: Note that its distance from the wing contour is constant initially and only grows when the flow slows down, indicated by the widening of the water jet. In contrast to this, on the photo in your question the flow moves away from the wing while still moving fast, indicating flow separation.

By the way, nice slats on that photo, it would be great if the computers figured out how much slat should be dropped for a given weight to optimize the wing drag (I understand they even tilt the lift vector forward a bit).

Will you please check this video? It is an wind tunnel test for varying AoA. In 5:44, it is clearly shown that the smoke is also deflected upward even with shallow AoA. The wind in the tunnel also is not high speed.

@AirCraftLover Now I wasted 8 mins watching this useless video. So what?

Sorry for your time you wasted. :)

As per my understanding from the video, the smoke in the tunnel was not high speed close to sound speed. And it was not wet wind. It just smoke, which flows quite slow. As in the 5:44, it was still deflected upward the smoke even it was low speed and shallow AoA. Then my question was, "What is the explanation of this airfoil phenomenon so the air from below the leading edge's tip deflected upward, even the wind flowed quite slow and with shallow AoA?" I rewrite the question to make better.

@AirCraftLover Speed does not matter much here. It is angle of attack in combination with camber. A small angle is enough for a significant upward deflection (also called induced angle of attack). This is absolutely ordinary and the mechanism is explained in lots of answers here.


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From Aerodynamic view, what is the advantage/disadvantage of the Akagi Kaman K-Max Helicopter?

Here is a video showing the helicopter, the rotors are perfectly synchronized.
As there are not many such design helicopter, then my question are, is any aerodynamic weakness? What is its advantage/disadvantage of that helicopter?

It's a twin rotor design, counter-rotating: no resulting torque on the airframe and no tail rotor needing to be powered. The twin rotors do not share a common shaft, with all of the extremely complicated control mechanisms that these require.
From J. Gordon Leishman, Principles of Helicopter Aerodynamics, section 2.15:

The design advantage of a twin rotor is that all power can be devoted to providing useful vertical lift and performance. However, the two rotors and their wakes interact with one another, producing a somewhat more complicated flow field than is found with a single rotor, and this interacting flow incurs a loss of net rotor system aerodynamic efficiency.

I believe they are very different between V-22 Ofsprey, Akagi Kaman K-Max, but probably a bit similar too Chinook Helicopter (also built by Boeing). What my concern/question here is due to the overlapping blades, which aerodynamically will disturb the wind. V-22's blades are not overlapped at all.

Indeed, because there is a wing in between the V22 main rotors, creating a much wider aircraft. If the V22 could not tilt its rotors forward, the design would not make much sense.


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Detection of solar balloons by ATC primary surveillance radar
I am curious whether a 30 cubic meter solar balloon made of black nylon and filled with ambient air would be detectable by the primary radar used by ATC in Continental Europe. Due to its size and light weight, a balloon of this kind is not required to carry a transponder to fly in European airspace. I wonder, given contemporary target-detection filters in primary radar relative to background 'clutter' or noise, whether an aerostat of this kind would be detectable.
Thank in advance,
Sasha

If it's filled with ambient air, how's it going to get off the ground?

@Harper: Black plastic absorbs sunlight, heats up, heats up the air inside, air expands, density becomes lighter than surrounding air, ascends.

#learnsomethingneweveryday @Nyos

This is not the same question as the lawn chair balloon question.  First, the lawn chair has a non-zero and significant RCS. Second, the man seated in the lawn chair has an even more significant RCS, and is detectable frequently by primary radar.  Third, a simple black plastic balloon has a very very small RCS, and would not be detected by primary radars.  So to answer the OP question, the key lies in the material used in the solar balloon.  It is likely that there will be a sputtered metalic coating, but if not the balloon in this question would have a slim chance of being detected.

@mongo, but the answer there is applicable also here (and what you say makes even more applicable). That makes it a duplicate

@Fredrico, would you not agree, since they are different, but related questions, closing this question as a duplicate does not give ample opportunity for answers which are specific to this different, but related, question?


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Do ducted fans have a future in forward flight aircraft? Can they be made more efficient than open rotors?
I'm fascinated by how ducted fans have the ability to be more efficient than open propellers of equal diameter in vertical flight through the use of tight tolerances, lightweight duct material, proper duct lip shape, and other factors. However, when learning that these benefits don't translate over into horizontal flight due to the duct's induced drag and pitching moment, I'm wondering if we'll ever see ducted fans in future aircraft.
I'm aware of the NASA ducted fan manned-aircraft project that failed in the 60's, but today I found out about AgustaWestland's Project Zero and noticed that they tilt the ducts and include fixed wings to provide sufficient lift. How would an aircraft like this compare efficiency-wise to something like a similarly sized  two rotor helicopter?
AgustaWestland Project Zero

Isn't this just a ultra high bypass turbofan without the turbo(turbine)?

@user3528438, not really; there is still a big difference between a short shroud like this and a long duct of a turbofan.

Fan shrouds are a fix for a problem that can normally be designed around in other, more efficient ways. They do reduce tip loses, but usually the tip loses are smaller than the weight and drag penalty of the shroud.
The fundamentals of prop/fan/rotor design is that thrust is proportional to the increased momentum of the air flow, but the power required to get that thrust is proportional to the added kinetic energy i.e. the ''square'' of the speed increase.
Therefore small, highly loaded fans will always be inefficient compared to a traditional helicopter rotor. The fact that they suffer from large tip losses only makes things worse.
So no, we won't see lots more ducts on future aircraft - except for on  jets, which use ducts because the size of their fans are limited by ground clearance and mitigating the losses from shock-waves when the fan blade tips go faster than sound.

Note that turbofans generally don't keep their fan blades subsonic. The tip speed goes up to around M1.5. The real reason for the duct is that the air exchanges speed for pressure in the inlet and flows through the duct at about the same speed independent of speed of the aircraft, so the fixed blades can do adequate work over large speed range. But it requires not just shroud, but duct of some minimal length.

Would the tip speed be kept subsonic relative to the air around the fan, whose velocity would be significant but subsonic relative to the shroud, or would the tip speed be supersonic within the surrounding air?

@JanHudec - thanks, I've corrected myself. I saw a good video on supersonic compressor blade tips recently but annoyingly can't find it now!

@supercat - air in the engine is usually referred to as 'slowed down' relative to the air passing the aircraft (i.e. from the aircraft's point of reference). Of course, from the ground, the air is accelerated. Either way, the fan blades are moving at right angles to the aircraft's motion, and the fan-blade tips are supersonic relative to the rest of the aircraft. What matters is that there are shockwaves inside the compressor.

@RobinBennett: Would the perpendicular component of the velocity of any part of the blade surface, other than a narrow leading edge, be anywhere close to supersonic, or would the angle of attack near the leading edge be shallow enough, and the curve designed in such a way that the perpendicular component would always be subsonic?

@supercat - I'm not sure what you're asking. Why do you think the leading edge is moving faster than the rest of the blade? What does angle of attack have to do with speed? The key here is that shockwaves occur when there's a change of pressure in supersonic flow, because that change of pressure can't propagate upstream.

@RobinBennett: Some portion of the leading edge of any practical blade will be perpendicular to the airflow, but if one could have an infinitely thin  blade and airflow were perfectly laminar and frictionless, a suitably-shaped blade of any size could travel through still air in an axial direction at any speed without disturbing any of the air at all, much less creating a shock wave.

@supercat - Ah, I see. However the point of a fan blade is to change the pressure of the air around it, so it's at an angle to the airflow, and cambered. Even if there's no stagnation at the front, the low pressure area on the front of the blade, and the high pressure area on the back are moving through the air.

@RobinBennett: Of course, but the velocity of for every area of the fan blade can be decomposed into a vector component which is perpendicular to its surface and a component which is coplanar with the blade, and in the absence of friction only the perpendicular component would affect the air.  Any portion of the surface whose perpendicular component of motion exceeds mach 1 would operate extremely efficiently, but I think spans with super-sonic tip speeds are designed to minimize those areas.


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How does corrugated skin influence the effectiveness of ailerons?
Recently I found a paper about incorporating corrugated skins in aircraft called 'Corrugated composite structures for aircraft morphing skin applications' by C. Thill, J.A. Etches, I.P. Bond, K.D. Potter and P.M. Weaver
Source: Paper PDF
The paper talks about how the application of corrugated skins might provide benefits in morphing skins. The paper mostly talks about the structural benefits, as requiring less force to change the shape. Even if the structure might be covered with an aerodynamic surface, I'm still interested in the aerodynamic properties of a corrugated surface.
If there's a wing surface with a corrugating perpendicular to the flow, as shown below (flow coming from the left).

Image taken from Paper PDF
How will such a surface incluence the effectiveness of non-corrugated ailerons?
I image such a surface would lead to a much thicker boundary layer, thereby greatly reducing aileron effectiveness.
I'm mostly interested in large corrugations(~50mm) and low speeds(~80km/h). I'm not asking about the lift generation of the surface itself, only the effect on a non-corrugated aileron.

As I read the paper, the corrugated element is covered by an elastic membrane that serves as the actual surface skin.

It mentions: "It is also planned to consider the aerodynamic consequences of having a corrugated wing profile at various Reynolds numbers." so it seems they are having some corrugated surface.

Sighardt Hörner to the rescue: Look here for data on the drag of grooves orthogonal to the flow direction. This is a non-starter with the dimensions and speeds you think about. Only covering this with a flexible skin will make this tolerable.

@PeterKämpf Thanks for the pointer. However, Hoerner talks mostly about the drag development, whereas I'm more interested in the flow development. What kind of flow structures can I expect, and how will these influence anything that's behind the corrugated surface?

@ROIMaison I'm speculating here, but expect oscillating separations (like von Karman streets) at the trailing edge. This will give you non-centering control surfaces with oscillating forces. No problem for hydraulic actuation, but not advisable for manual control. Still, flutter can be expected in either case.

Yes the skin will be covered by elastomeric material to maintain the aerodynamic profile. However, someone studied the aerodynamics of corrugated panels. Please check this paper: https://arc.aiaa.org/doi/abs/10.2514/6.2014-2724


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What is difference between pitch & lift?
What is the difference between pitch and lift of an aircraft?Please kindly try to include graphics with your answers

Too elementary, no evidence of research effort, suggests no understanding of the vocabulary at all.  Plus it it essentially (probably unintended) trolling for someone to offer all kinds of inaccurate answers, like "pitch controls lift" or "lift controls climb/sink rate controls pitch", etc.

The pitch axis is one of the three axes around which an aircraft can rotate. The pitch angle is the angle of rotation around this axis, where zero indicates that the aircraft is level with the horizon (in pitch). You can see all three axes in the following image (from Wikimedia):

See also What are the exact meanings of roll, pitch and yaw?
All three angles combined describe the attitude of the aircraft. The term pitch attitude is used to describe the current pitch angle.
The lift is a force acting on the aircraft. It is mostly created by the wings, but also by the horizontal stabilizer and less so by the rest of the aircraft. You can see the typical 4 forces acting on an aircraft in the following image (from NASA):

In level flight the lift is exactly equal to the weight.

The pitch angle and lift have no direct connection. Lift does indirectly depend on pitch because increasing pitch will increase the angle of attack, which will influence lift. See also What effect does the angle of attack have on lift and drag?

Answer could be altered / expanded to include term "pitch attitude" as that seems a useful thing to include in an explanation of basic concepts.  E.g. "The pitch angle or pitch attitude is....".

@quietflyer OK, included a reference for attitude in general and mentioned pitch attitude in particular.


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Confusion regarding helicopter?
Is there any difference for the case when helicopter rotor spins clockwise and the case when helicopter rotor spins anticlockwise? Is it something like clockwise spinning will cause the helicopter to move up and to move down for anti clock wise?

Helicopter blades always produce downward thrust. The amount of thrust will determine whether the helicopter is moving up, or down, or hovering stationary. The amount of thrust is controlled by blade pitch, not direction of rotation.

This might also be interesting.

Rotors only spin one way at all times, like airplane propellers, because thrust is adjusted with blade pitch (even with a propeller that has reverse thrust, it's done by reversing the pitch, not by driving the prop in the other direction).
But you'll find rotors spin in different directions depend on manufacturer. Generally in NA helicopters are counterclockwise looking down. Some European machines spin the opposite way.
The biggest practical effect is the direction of rotation determines the left/right orientation of certain tail rotor control characteristics, because the tail rotor is making its thrust blowing air left or right depending on the main rotor's torque direction.
More importantly, it determines the direction to roll if you are ditching the thing in water, to make sure the blades contact the water behind you. If you have to set a land based Bell 206 on the water (assuming no airbag safety system so you are going on your side anyway), you initiate a roll to the right at touchdown toward the advancing blade, so that the first blade that hits the water hard enough to stop it does so to the right of the machine and the following blades slam into the water more or less over the tail boom. Roll the other way so the blades contact the water left side first, the blades following the first one to stop can come into the cockpit roof.
If your machine's rotor spins clockwise from above, like most Eurocopter machines, you roll it to the left to get the same effect.

Helicopter rotors are designed to rotate only in one direction. But the designer can freely choose the direction, and both clockwise and counter-clockwise rotors are commonly found in service today.
The rotor torque and some aerodynamic effects are based on the direction of rotation, and so these effects can be in opposite directions between a clockwise and a counter-clockwise rotating helicopter. But "up to symmetry" the effects are identical.
Thrust is adjusted by varying the pitch of the rotor blades, not by changing the speed or the direction of rotation.


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How is pitch changed and lift generated in QNET VTOL (Vertical Take off & Landing)?
I am confused how QNET VTOL (Vertical Take off & Landing) works, especially how pitch is changed and lift is generated.
I share a link of a relevant YouTube video and I am also sharing snaps of the top view and front side view of this trainer.
I am confused regarding a very basic question but I am not able to find its answer: according to my understanding here in this VTOL (Vertical Take off & Landing) trainer, a fan will act like the rotor of a helicopter (based upon the Bernoulli principle), and when this fan rotates with significant speed because of significant current provided to its motor, lift will be generated.
Please correct me if my understanding is not correct.
Top view

Front side view

Related: https://electronics.stackexchange.com/questions/464317/confusion-regarding-qnet-vtolvertical-take-off-and-landing

Being migrated to Aviation

You got itexactly right: Current can be applied to the Motor which starts spinning. Attached to the Motor is an airscrew (Propeller), which turns when the Motor turns. This Propeller generates a force which then pushes on the beam hence moving it up or down, exactly as you pointed out. Where lies the confusion?

Also there may be a confusion here with use of 'pitch'. In the related question pitch refers to the angle of the beam holding the motor and fan. Not pitch of the blades like a true helicopter. There are only two wires going to the motor so it is a very simple cast plastic fan in the housing. So the lift is only controlled by the motor speed - recommend not over thinking this.

This device might be useful for teaching engineering concepts, but the "Verti-bird" I had as a kid was a better VTOL trainer:  https://www.youtube.com/watch?v=FojzXSa6bxk

Your understanding is correct.

The motor is controlled by the input current.
When the motor and the attached propeller spins, lift is generated.
The lift counteracts the gravity and lifts the beam up.
The pitch angle of the beam is measured and used as an input to the control system.

Note that this is  different from a helicopter: the primary means of changing the lift component in a helicopter is to collectively change the angle of attack of the rotor blades. The rotor's rotational speed is kept approximately constant.

In such simple toys, it is not the pitch of the blades to be changed to control thrust rather the rotating speed of the fan via the power supplied to/from the electrical engine. Changing the pitch would be way to complicated and delicate (source):

The simple momentum theory allows us to go from the power generated by the engine to the thrust (thrust, not lift) generated by a rotor/propeller:
$ T = \sqrt[3]{2 \rho A P^2} $
where $A$ is the rotor area, $\rho$ is the density and $P$ the power needed to spin the rotor.
For an electrical engine the power is simply $P=V\cdot I$ times an efficiency factor $\eta$.
The fan in the picture is ducted so that we can assume some 30% thrust increment the power being the same.
The thrust equals and is opposite to the weight of the arm+propeller assembly.
Knowing engine ratings, weights and disk area it is possible to do the calculation. If you know these values I might complete my answer with real numbers.


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What is the major difference between this VTOL setup and an actual helicopter, in terms of degree of freedom?
I am studying QNET VTOL(Vertical Take off & Landing) system. I want to know how it is different from an actual helicopter . According to my understanding, an actual helicopter has 6 degrees of freedom while this VTOL system has only 1 degree of freedom as it pitch is the only parameter here that is variable.I am also attaching snapshots of top & front  view of this system and also i am giving below link of a relevant youtube video.

Please correct me if my understanding is not correct

I guess it is controlled by some other microcontroller but i am pretty much sure that aurdino is not being used here. It is something else

Ok, thanks for the answer :)

The differences are endless, the only thing in common is the name VTOL
A helicopter has indeed 6 degrees of freedom. The helicopter is controllable by 5 input variables: lateral & longitudinal cyclic, collective, anti-torque pedals and power). But non of the inputs correlate purely with a single degree of freedom; there is a lot of cross talking between the control variables.
Change any one of the inputs and you disturb the balance of forces and moments so that you have the change something else as well.
For example, if you change the collective to climb, you will also require more power to avoid the rotor from slowing down. More power means more torque on the rotor, so now you also need to control the anti-torque pedals to balance that out. But the tail rotor will then change the side force on the helicopter, so now you need lateral cyclic to compensate for that. That upsets the vertical equilibrium slightly and so we are back at the start.
Another major difference is the stability. Helicopters need constant control inputs to stop them from diverging from the initial state. The QNET VTOL would very likely find a balance point for a given control input setting. A helicopter, being unstable, would diverge from its initial state and crash, not matter what constant input you would give. It needs constant input adjustments to keep it in check.
From a control system point of view, this QNET VTOL system seems to be a nice tool to get practical experience with modeling, system parameter identification and PID control. However there is still an enormous step to go before one can design a helicopter control system. An intermediate step would be controlling a quad-copter which is in terms of control system complexity somewhere in the middle (but still very far from the QNET VTOL)

A key difference here is stability - I would expect a fixed power level would damp to a fixed position even without electronic assistance on this due the presence of a ground effect. VTOL aircraft are infamously unstable so poor control system setup will diverge and crash quickly. If the setup allowed the arm to go vertical that might be a closer example of flight control.

@GremlinWranger good point. I'll add a paragraph about stability

"...the only thing in common is the name VTOL."  Which is a misnomer because that machine is neither V, nor TO, nor L.

@MichaelHall absolutely agree. There is nothing that this thing and a helicopter have in common.

What is the major difference in terms of degree of freedom?

In terms of degrees of freedom (dof) a helicopter's fuselage has indeed 6 dof but then you have to add the dof of the blades (each of them pitches, lead-legs and flaps) and the dof of the tail rotor (pitch of its blades).
That if you consider the structure as being rigid, otherwise the dof due to the flexibility should be considered as well... helicopters are quite complicated :-)
By extension also your experiment has actually 2 dof: its angle in respect to the hinge and the speed of the propeller which, I suppose, is controlled via the voltage applied to the electrical motor driving the propeller.